A compound (MO_2) of a group 1 element (M) is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The hydrolysis of a superoxide $(MO_2)$ of a group $1$ element $(M)$ is given by the reaction: $2MO_2 + 2H_2O \longrightarrow 2M^{+} + 2OH^{-} + H

Vedclass · Accepted Answer

$H_2O_2, O_2$

Vedclass · Answer

(A) The hydrolysis of a superoxide $(MO_2)$ of a group $1$ element $(M)$ is given by the reaction: $2MO_2 + 2H_2O \longrightarrow 2M^{+} + 2OH^{-} + H_2O_2 + O_2$. Here,$X$ is $H_2O_2$ and $Y$ is $O_2$. When $X$ $(H_2O_2)$ reacts with $I_2$ in a basic medium,the reaction is: $I_2 + H_2O_2 + 2OH^{-} \longrightarrow 2I^{-} + 2H_2O + O_2$. In this reaction,$I_2$ is reduced to $I^{-}$ and $H_2O_2$ is oxidized to $O_2$ $(Y)$. Thus,$X$ is $H_2O_2$ and $Y$ is $O_2$.

$A$ compound $(MO_2)$ of a group $1$ element $(M)$ is hydrolyzed to form $M^{+}$,$OH^{-}$,$X$,and $Y$. When $X$ reacts with $I_2$ in a basic medium,the products formed are $I^{-}$,water,and $Y$. Then $X$ and $Y$ are respectively:

Similar Questions

$H_2O_2$ reduces $K_4[Fe(CN)_6]$ in which of the following conditions?

The volume concentration of hydrogen peroxide having $6.8 \%$ concentration will be

The compound that can work both as an oxidising and reducing agent is

The amount of $H_2O_2$ present in $1 \ L$ of $1.5 \ N$ $H_2O_2$ solution is ........... $g$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module